# What is an Intuitive Explanation of the Phase of a Signal

I understand that the meaning of the phase response of a system is simply how much the system delays a frequency component. However, I do not find an intuitive explanation for the phase of a signal. The representation the Fourier transform of a signal by polar coordinates, produces two components; the magnitude and the phase. The magnitude response is simply how much each frequency component is contributing to this signal. However, what is the meaning of the phase here? Can I know if a frequency component is "exist" in the signal by just looking into the phase of a signal?

I found an explanation here:

determines how the sinusoids line up relative to one another to form a signal

But the did not help me.

Compare $y= sin(\omega t)$ to $y=sin(\omega t-\phi)$ as shown in the plot below. $\phi$ is the additional phase term in radians where $\omega$ represents the frequency in radians per second. Thus the phase term shifts a sinusoid along the horizontal axis. So at a given frequency this will result in a time delay for that frequency, although to be noted that a fixed time delay will cause a phase shift that is proportional to frequency as a linear variation (The phase will increase linearly as the frequency increases). Thus our interest in "linear-phase" filters as this will have a constant "group delay" meaning the group of all frequencies in a waveform will be delayed the same amount in time, and thus not cause destructive interference between the different frequencies in the group. (Review Fourier Series and decomposition of waveforms into individual frequencies and the resulting reconstruction for more insight on that).

More pertinent (and useful) is complex signal representation with $Ae^{j \phi}$ representing the magnitude A and phase $\phi$ of a signal. This can be expanded using Euler's identity which may help give further insight: $$Ae^{j\phi}=Acos(\phi)+jAsin(\phi) = I + jQ$$

From which we see that in order to completely describe a complex signal we require two real signals, this could be magnitude and phase, A, $\phi$ as in $Ae^{j\phi}$ or the real and imagninary components I, Q as in I+jQ.

Refer to this post which may help give further insight into the importance and utility of complex representation for signals: Frequency shifting of a quadrature mixed signal

• I understand from your answer that the phase response of a signal is simply, when a frequency shows up in the signal. For example, on t = 5 seconds, a 5 Hz frequency will appear. am I correct? – hbak May 11 '17 at 15:04
• No other lines are not necessarily "phase response" but phase of the signal itself. The signal can be represented as a real sine wave or as a complex signal. As described, the frequency is always there, it does not "appear" in 5 seconds. The frequency describes the rate of change of phase. And the phase shows the relative magnitudes of the signal on the real and imaginary axis, which is changing with time sinusoidally for a given fixed frequency. – Dan Boschen May 11 '17 at 15:08
• "The phase shows the relative magnitudes of the signal on the real and imaginary axis, which is changing with time sinusoidally for a given fixed frequency.". Can you simplify that more? – hbak May 11 '17 at 15:19
• @hbak I suggest you do the following excercises: plot the function $cos(\omega t)$ for a fixed constant $\omega$ and varying t, use $\omega = 20$ and vary t from 0 to 1, then make another plot where you plot the same $cos(\omega t)$ on the x axis vs $sin(\omega t)$ on the y axis and tell me what you get for a result. – Dan Boschen May 11 '17 at 17:36
• I attached the image to my question. – hbak May 11 '17 at 19:24

The phase for each bin in the FFT is the same as the relative phase shift of the sinusoid that represents that bin in the time domain (and its complex conjugate symmetric pair to be a real sinewave). That phase is the starting phase in time for that component such that when added to all the other sinewaves from the other bins in frequency will properly add up to recreate the signal in time.

The phase of a signal tells one nothing without the magnitude. FFT result bins within a rounding error of zero often have random phases. Whereas the angle of a non-zero length vector actually points somewhere.

Note that a cosine and a sine of the same frequency are orthogonal. A ratio of the two (cosine and sine) is required to represent any sinusoid of that same frequency. The phase tells you the ratio.

• The phase of a complex baseband BPSK signal tells you everything you need to know about the signal from a demodulation perspective – BigBrownBear00 May 12 '17 at 1:31
• Not if the magnitude is zero or otherwise just noise (depending on the coding gain). – hotpaw2 May 12 '17 at 2:43
• yes, even if the magnitude is zero or just noise. – BigBrownBear00 Feb 1 '18 at 20:05

The phase of a signal generally refers to the timing of the signal (or how two sinusoids line up) as you posted in your question. But you are asking about the phase of a signal in the frequency domain (i.e., after an FFT operation). The FFT function computes an N-point complex DFT. In practice the real and imaginary components of the result of the FFT operation are used to generate the magnitude spectrum which is usually more useful than the individual I and Q components. The magnitude of the FFT can be used for spectral visualization, energy detection, or a variety of other applications.

• The phase of the frequency components can also be very useful/important. For example, accessing stability of control systems with Bode plots in the frequency domain. – user27621 Feb 1 '18 at 16:34

Michael Ossman likens the phase of a signal to a slinky (see video here). Instead of thinking of a wave as 2-dimensional, imagine a 3-dimensional helix (a Slinky), where phase is the deviation from the carrier signal.